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Math Help - limit definition of derivative

  1. #1
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    limit definition of derivative

    3)Use the limit definition of derivative and find, y'
    for \frac{1}{x+1}

    I know have the answer because we are taught the short cut method in school where we multiply the coefficient by the power and decrease by one. But I want to see how this is done the longer method.
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  2. #2
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    Find this limit.
    \displaystyle\lim _{h \to 0} \frac{{\frac{1}<br />
{{x + 1 + h}} - \frac{1}<br />
{{x + 1}}}}<br />
{h}
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  3. #3
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    I end up with:

    \frac{h^2}{(x+1)^2 +h(x+1)}
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  4. #4
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    Quote Originally Posted by elieh View Post
    I end up with: \frac{h^2}{(x+1)^2 +h(x+1)}
    You made some algebra mistakes.
    \dfrac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h}=\dfrac{\frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}}{h}
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  5. #5
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    \frac{-h^2}{(x+1)^2 + h(x+1)}

    If h is on top and it tends to 0 then the limit is 0...
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  6. #6
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    The answer is not zero.
    Look at post #4. Can you do the algebra?
    You said that you already know the answer: \dfrac{-1}{(x+1)^2}
    So work towards that answer.
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  7. #7
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    Got it thanks!
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